When does a square matrix is equivalent to the identity matrix? Why does a square matrix A with an order of $n \times n$ and $rank(A)=n$ can be reduced to Identity matrix (or by multiplying a sequence of elementary matrices)
 A: the rank of $A$ is the number of linearly independent columns of $A$ or, in other words, the dimension of the image of $A$ as a vector space.
Your $n\times n$ matrix $A$ corresponds to a linear transformation $T:R^n \to R^n$, so for it to have rank $n$ it means it is a linear isomorphism. 
This means the vectors $T(e_i)$ for $i=1,\dots n$ form a basis for the codomain. If you write $T$ in these coordinates, the matrix $A'$ you get is the identity. 
A: You must know something about elementary operations and rank in order to understand this property. I am going to suppose you know that row operations used during Gaussian elimination amount to left-multiplying by elementary matrices, and that these operations do not change the linear dependence of any set of columns; the rank is the size of a maximal set of linearly independent column, so the rank does not change under row operations.
Now imagine a matrix that cannot be transformed to the identity matrix using row operations. This means that at some point one cannot choose a pivot in column$~j$ for some $1\leq j\leq n$. If the pivots in any previous columns have been moved to rows${}<j$ this means that at this point column$~j$ has zero entries in rows $j,\ldots,n$, and by the way Gaussian elimination operates (the previous columns also have zero entries in those rows) this means column$~j$ is linearly dependent with the previous columns. By the invariance mentioned above, this linear dependence is also present in the corresponding columns of the original matrix. Thus the set of all columns is not linearly independent, and the rank cannot be$~n$.
A: The magic words are "Gaussian elimination".
